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gavmur [86]
3 years ago
8

On July 23, 1983, Air Canada Flight 143 required 22,300 kg of jet fuel to fly from Montreal to Edmonton. The density of jet fuel

is 0.803 g/mL, or 1.77 lb/L. The plane had 7682 L of fuel on board in Montreal. The ground crew there multiplied the 7682 L by the factor 1.77(without units) and concluded that they had 13,597 kg of fuel on board and needed an additional 8703 kg by the factor 1.77(again without units) and concluded they needed to add 4916 L of fuel. They added 5000L. On its flight, the plane ran out of fuel and crash-landed near Winnipeg, hundreds of kilometers short of its destination. (there were few injuries and no fatalities.) What mistake did the ground crew make? How much fuel(in L) should they have added before takeoff?
Engineering
1 answer:
Natasha2012 [34]3 years ago
5 0

Answer:

20, 083 L

Explanation:

The mistake was the result of not using units when converting the 7862 l to Kg. They used the density in pounds hence they multiplied by 1.77 Lb/L and obtained 13597 Lb not Kg as they assumed.

To obtain the amount needed to refuel they subtracted this quantity from the 22,300 Kg required for the trip again obtaining the wrong quantity of 8703 Kg and they converted this to liters by dividing the density to get 4916 L and then placed then 5000 L of fuel

The quantity required was

7862 L * 1.77 Lb/L = 13915.74 Lb (pounds not kilos)

then converting this pounds to Kg by multiplying by  0.454 Kg/L one gets

6173 Kg on board

Amount Required

( 22,300 -6173)  :  16127 Kg

16127 Kg/ 0.803 Kg/L =  20083 L

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An ideal reheat Rankine cycle with water as the working fluid operates the boiler at 15,000 kPa, the reheater at 2000 kPa, and t
solniwko [45]

Answer:

See the explanation below.

Explanation:

First find the enthalpies h₁, h₂, h₃, h₄, h₅, and h₆.

Find h₁:

Using Saturated Water Table and given pressure p₁ = 100 kPa

h₁ = 417.5 kJ/kg

Find h₂:

In order to find h₂, add the w_{p} to h₁, where  w_{p}  is the work done by pump and h₁ is the enthalpy computed above h₁ = 417.5 kJ/kg.

But first we need to compute  w_{p} To computer  

Pressures:

p₁ = 100 kPa

p₂ = 15,000 kPa

and

Using saturated water pressure table, the volume of water v_{f} = 1.0432

Dividing 1.0432/1000 gives us:

Volume of water = v₁ =  0.001043 m³/kg

Compute the value of h₂:

h₂ = h₁ + v₁ (p₂ - p₁)

    = 417.5 kJ/kg + 0.001043 m³/kg ( 15,000 kPa - 100 kPa)

    =  417.5 + 0.001043 (14900)

    = 417.5 + 15.5407

    = 433.04 kJ/kg

Find h₃  

Using steam table:

At pressure p₃ = 15000 kPa

and Temperature = T₃ = 450°C

Then h₃ = 3159 kJ/kg

The entropy s₃ = 6.14 kJ/ kg K

Find h₄

Since entropy s₃ is equal to s₄ So

s₄ = 6.14 kJ/kgK

To compute h₄

s₄ = s_{f} + x_{4} s_{fg}

x_{4} = s_{4} -s_{f} /s_{fg}

x_{4} = 6.14 -  2.45 / 3.89

x_{4}   = 0.9497

The enthalpy h₄:

h₄ = h_{f} +x_{4} h_{fg}

    = 908.4 + 0.9497(1889.8)

    =  908.4 + 1794.7430

    = 2703 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₃ = 6.14 kJ/ kg K and pressure p₄ = 2000 kPa

Find h₅

Using steam table:

At pressure p₅ = 2000 kPa

and Temperature = T₅ = 450°C

Then h₅  = 3358 kJ/kg

Find h₆:

Since the entropy s₅ = 7.286 kJ/kgK is equal s₆ to  So

s₆ = 7.286 kJ/kgK = 7.29 kJ/kgK

To compute h₆

s₆ = s_{f} + x_{6} s_{fg}

x_{6} = s_{6} -s_{f} /s_{fg}

x_{6} = 7.29 - 1.3028 / 6.0562

x_{6}   = 0.988

The enthalpy h₆:

h₆ = h_{f} +x_{6} h_{fg}

    = 417.51 + 0.988 (2257.5)

    = 417.51 + 2230.41

  h₆ =  2648 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₅ = 7.286 kJ/kgK and pressure p₅ = 2000 kPa

Compute power used by pump:

P_{p} is found by using:

mass flow rate = m =  1.74 kg/s

Volume of water = v₁ =  0.001043 m³/kg

p₁ = 100 kPa

p₂ = 15,000 kPa

P_{p}  = ( m ) ( v₁ ) ( p₂ - p₁ )

     = (1.74 kg/s) (0.001043 m³/kg) (15,000 kPa - 100 kPa)

     = (1.74 kg/s) (0.001043 m³/kg) (14900)

     = 27.04

P_{p} = 27 kW

Compute heat added q_{a} and heat rejected q_{r}  from boiler using computed enthalpies:

q_{a} = ( h₃ - h₂ ) + ( h₅ - h₄ )

      = ( 3159 kJ/kg - 433.04 kJ/kg ) + ( 3358 kJ/kg - 2703 kJ/kg )

      = 2726 + 655

      = 3381  kJ/kg

q_{r} =  h₆ - h₁

  = 2648 kJ/kg - 417.5 kJ/kg

  = 2232 kJ/kg

Compute net work

W_{net} = q_{a} - q_{r}

       = 3381  kJ/kg - 2232 kJ/kg

       = 1150 kJ/kg

Compute power produced by the cycle

mass flow rate = m =  1.74 kg/s

W_{net} = 1150 kJ/kg

P = m * W_{net}

  = 1.74 kg/s * 1150 kJ/kg

  = 2001 kW

Compute rate of heat transfer in the reheater

Q = m * ( h₅ - h₄ )

   =  1.74 kg/s * 655

   =  1140 kW

Compute Thermal efficiency of this system

μ_{t} = 1 - q_{r} /  q_{a}

   = 1 - 2232 kJ/kg / 3381  kJ/kg

   = 1 - 0.6601

   = 0.34

   = 34%

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